Problem: The equation of a circle $C$ is $x^2+y^2-12x-16y+96 = 0$. What is its center $(h, k)$ and its radius $r$ ?
To find the equation in standard form, complete the square. $(x^2-12x) + (y^2-16y) = -96$ $(x^2-12x+36) + (y^2-16y+64) = -96 + 36 + 64$ $(x-6)^{2} + (y-8)^{2} = 4 = 2^2$ Thus, $(h, k) = (6, 8)$ and $r = 2$.